X > 0, but why doesn't 1/x > 0?

[FredericChopin]

If a > b, where a > 0 and b > 0, then 1/a < 1/b.

But what if b = 0? For example, if x > 0, meaning if x is a positive number, then it should be that 1/x > 0.

Yes, yes, I know I would be dividing by 0, but it doesn't make sense intuitively. If x is a positive number, then obviously 1/x is a positive number. So it should be that 1/x > 0.

Can somebody explain what is going on here?

Thank you.

 

[Simon Bridge]

You have the answer yourself, 1/0 is "undefined" for a reason.

but it doesn't make sense intuitively

... well 1/0 makes no sense at all. If you intuitively expect it to then your intuition is wrong.

To make sense of things, you have to ask, instead, what happens to 1/x when x approaches zero trough positive reals... i.e. for that case that b is arbitrarily small.
Try that.

 

[FredericChopin]

If x approaches 0 from the right, then 1/x will become arbitrarily (positively) large, which is definitely still greater than 0. Hm... I still don't see what could be wrong.

Before I posted, I Googled "inequality properties" and I found a popular school study guide called "Sparknotes", which has a reputation for being inaccurate. However, on their page on inequalities:

http://www.sparknotes.com/math/algebra1/inequalities/section2.rhtml

, they mention that if a > 0, then 1/a > 0 (see the attached screenshot from the page). Is this wrong too? Well, it wouldn't be a surprise if it was.

Thank you.

 

[SteamKing]

If x approaches 0 from the right, then 1/x will become arbitrarily (positively) large, which is definitely still greater than 0. Hm... I still don't see what could be wrong.

Before I posted, I Googled "inequality properties" and I found a popular school study guide called "Sparknotes", which has a reputation for being inaccurate. However, on their page on inequalities:

http://www.sparknotes.com/math/algebra1/inequalities/section2.rhtml

, they mention that if a > 0, then 1/a > 0 (see the attached screenshot from the page). Is this wrong too? Well, it wouldn't be a surprise if it was.

Thank you.

Well, run a test. If a = 2, then is a > 0? What about 1/2? Is 1/2 > 0? These are trivial cases which show that if a > 0, then 1/a > 0.

 

[symbolipoint]

Approaching a value and equality to a value are different things.

 

[Simon Bridge]

if x approaches 0 from the right, then 1/x will become arbitrarily (positively) large, which is definitely still greater than 0. Hm... I still don't see what could be wrong.

There's nothing wrong with that statement.

[sparknotes] mention that if a > 0, then 1/a > 0 (see the attached screenshot from the page). Is this wrong too?

No. That is correct.
I don't see how you would have thought that may be wrong and it is simple to check as SteamKing suggests.

The rule you started with, 1/a < 1/b : a>b>0, simply does not apply for b=0, just like it does not apply for a=b or or a<b.
Further, "1/a > 0" only holds for a>0 and "1/a" is undefined for a=0.

So what is the problem?

 

[FredericChopin]

There's nothing wrong with that statement.

No. That is correct.
I don't see how you would have thought that may be wrong and it is simple to check as SteamKing suggests.

The rule you started with, 1/a < 1/b : a>b>0, simply does not apply for b=0, just like it does not apply for a=b or or a<b.
Further, "1/a > 0" only holds for a>0 and "1/a" is undefined for a=0.

So what is the problem?

Oh, well, there is nothing wrong, then. I thought that rule was wrong because if x > 0, then dividing 1 by both sides of the inequality yields 1/x > 1/0. My concern was that 1/0 is not defined. There is clearly something wrong with the algebra, but what is wrong?

Thank you.

 

[Simon Bridge]

I thought that rule was wrong because if x > 0, then dividing 1 by both sides of the inequality yields 1/x > 1/0 ... There is clearly something wrong with the algebra, but what is wrong?

If I follow you, you are asking if: $$\frac{1}{a>b} = \frac{1}{a} > \frac{1}{b}$$ is proper algebra?

 

[FredericChopin]

If I follow you, you are asking if: $$\frac{1}{a>b} = \frac{1}{a} > \frac{1}{b}$$ is proper algebra?

Rather, something like this:

If a ≠ 0 and b ≠ 0, then if a = b, then 1/a = 1/b.

Well, I'm probably not using the inequality sign correctly and trying to use equality rules for the "greater than" sign. But basically, if x > 0, then a "=" x, and b "=" 0, and using the algebra "rule" above, if x > 0, then 1/x > 1/0, which is incorrect.

 

[Simon Bridge]

Certainly a sequence of statements that is true for an equality need not be true for an inequality.

However, I am having trouble figuring out what you are talking about.

if x > 0, then dividing 1 by both sides of the inequality yields 1/x > 1/0

... this would be a description of some algebra - but a vague description.
"Divide 1 by both sides of the inequality" does not make mathematical sense, so I suspect this is where you have erred.
Can you show a series of algebraic steps that start with x>0 and end with 1/x > 1/0 ?
After all: if the former is true then the latter is false.

i.e. what is it that is wrong?

 

[soarce]

$a>b$, then multiply both sides by $1/ab$ (with $a,b>0$) and obtain $1/b>1/a$

$x>0$, then mutliply both sides by $1/x^2$ and obtain $1/x>0$

 

[Simon Bridge]

@FredericChopin: soarce's post is the kind of thing I was trying to get you to do.
Does that help?

 

[symbolipoint]

Why the continued confusion? Approaching and Equaling are not the same.